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grandymaker [24]

2 years ago

8

A pizza place offers 4 different cheeses and 10 different toppings. In how many ways can a pizza be made with 1 cheese and 4 top

pings?

1 answer:

mylen [45]2 years ago

8 0

Answer:

Step-by-step explanation:

Assuming the toppings don't have to be the same

4*10*10*10*10=4000

If the toppings have to be different:

4*10*9*8*7=20160

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The original rectangle has a perimeter of 16

Serjik [45]

12.4 / 6.2 = 2

SF = 2

The original rectangle has a perimeter of 16

Perimeter of the enlarged rectangle = 16 x 2 = 32

Answer: 32 feet

6 0

3 years ago

Read 2 more answers

(a^(2)-4ac+3bc)/(a^(2)-ab+bc-ac)+(a+3b)/(b-a)+(a+2c)/(a-c)

Bingel [31]

Answer:

The answer is −a^4 + 5a^3b + 2a^3c − 4a^2b^2 − 10a^2bc − a^2c^2 + 8ab^2c + 5abc^2 − 4b^2c^2 / −a^4 + 2a^3b + 2a^3c − a^2b^2 − 4a^2bc − a^2c^2 + 2ab^2c + 2abc^2 − b^2c^2

Step-by-step explanation:

This the equation being simplified

4 0

3 years ago

Read 2 more answers

the depth D, in inches, od wsnow in my yard t hours after it started snowing this morning is given by D=1.5t + 4. if the depth o

Blababa [14]

Answer:

8.5 inches

Step-by-step explanation:

First let's find the time t when the depth of the snow is 7 inches.

To do this, we just need to use the value of D = 7 then find the value of t:

$7 = 1.5t + 4$

$1.5t = 3$

$t = 2\ hours$

We want to find the depth of snow one hour from now, so we just need to use the value of t = 3 to calculate D:

$D = 1.5*3 + 4$

$D = 4.5 + 4 = 8.5\ inches$

The depth of snow one hour from now will be 8.5 inches.

7 0

3 years ago

Answer this question for me please cause idk it

Greeley [361]

H=-4.9t^2+25t 0=t(-4.9t+25) T=25÷4.9 T= approx 5.1 which is closes to 5.0 So answer will be 5.0

8 0

2 years ago

Suppose the horses in a large stable have a mean weight of 1467lbs, and a standard deviation of 93lbs. What is the probability t

krok68 [10]

Answer:

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 1467, \sigma = 93, n = 49, s = \frac{93}{\sqrt{49}} = 13.2857$

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable?

This is the pvalue of Z when X = 1467 + 9 = 1476 subtracted by the pvalue of Z when X = 1467 - 9 = 1458.

X = 1476

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{1476 - 1467}{13.2857}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

X = 1458

$Z = \frac{X - \mu}{s}$

$Z = \frac{1458 - 1467}{13.2857}$

$Z = -0.68$

$Z = -0.68$ has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the mean weight of the sample of horses would differ from the population mean by less than 9lbs if 49 horses are sampled at random from the stable

5 0

3 years ago